# force field work done

A force field in 3-space is given by the formula $F(x,y,z)=(x+yz,y+xz,x(y+1)+z^2)$. Calculate the work done by F in moving a particle once around the triangle with vertices $(0,0,0)$, $(1,1,1)$, $(1,1,0)$ in that order.

So $W=\int \limits_{C} Fdr$ but how do I make my $r(t)$ function? Do I do it part by part as in let A, B, C be the vertices of the triangle respectively with the order they are given. First find the vector equation of AB intersecting with A and B which would be say $r_1(t)= (0,0,0)+(1,1,1)t=(t,t,t)$ right? But this is just a guess. Even if this is right, what would the limits even be.

• Have you studied line integral? – Mhenni Benghorbal Jan 28 '15 at 13:32
• you're on the right track. over what values of $t$ does your $r_1(t)$ trace out $AB$? – Callus - Reinstate Monica Jan 28 '15 at 13:34
• $r_1 ^{'}(t)=(1,1,1)$ and $F(r_1 ^{'}(t))=(t+t^2, t++t^2, t+2t^2)$. So $$\int Fdr=\int F(r(t))r'(t)dt=\int \limits_C (4t^2+3t)dt=[\frac{4t^3}3+\frac{3t^2}2]_C$$ but I don't know the limits... – snowman Jan 28 '15 at 13:40
• Linear interpolation between $P$ and $Q$ can be achieved via $(1-t) P + t Q$ for $t \in [0,1]$. – mvw Jan 28 '15 at 13:43
• @Callus Can you have a look at what I traced out please. I don't know how to find the limits. – snowman Jan 28 '15 at 13:44

In a vector setting $$W=\int F\cdot\mathrm{d}r$$ We can parametrize the first side of the triangle by $r=(t,t,t)$ for $t$ from $0$ to $1$: \begin{align} W_1 &=\int_0^1(t+t^2,t+t^2,t+2t^2)\cdot(1,1,1)\,\mathrm{d}t\\ &=\int_0^1(3t+4t^2)\,\mathrm{d}t\\ &=\frac32+\frac43\\ &=\frac{17}6 \end{align} We can parametrize the second side of the triangle by $r=(1,1,1-t)$ for $t$ from $0$ to $1$: \begin{align} W_2 &=\int_0^1(2-t,2-t,2+(1-t)^2)\cdot(0,0,-1)\,\mathrm{d}t\\ &=\int_0^1(-2-(1-t)^2)\,\mathrm{d}t\\ &=-2-\frac13\\ &=-\frac73 \end{align} The third side is similar.

• @mvw: ah, thanks. I was remembering the denominator for $W_1$ – robjohn Jan 28 '15 at 18:14

One could as well evaluate it via the Kelvin-Stokes theorem: $$W = \int\limits_{\partial T} F \cdot du = \int\limits_{T} \mbox{rot } F \cdot dA$$ where $\mbox{rot } F = (\epsilon_{ijk}\partial_j F_k) = (0, -1, 0)^T$ and $dA = \frac{1}{\sqrt{2}}(-1,1,0)^T\,\lVert dA \rVert$ which gives the constant $-\frac{1}{\sqrt{2}}$ as integrand. The orientation of $T$ is given by the right hand rule.

That leaves the proper 2D integration over the triangle as calculating its area. It should have $\lVert A \rVert = \frac{\sqrt{2} \times 1}{2} = \frac{1}{\sqrt{2}}$

That should give $W = -\frac{1}{2}$ if I did not make a mistake.

Checking

Integrating along $u = (1, 1, 1-t)^T$ gives $du = (0, 0, -1)^T dt$ and \begin{align} W_2 &= \int\limits_0^1 (*,*,2+(1-t)^2)^T\cdot(0, 0, -1)^T dt \\ &= \int\limits_0^1 (-t^2 + 2t - 3)dt \\ &= \left[-\frac{1}{3} t^3 + t^2 - 3t\right]_0^1 \\ &= -\frac{7}{3} \end{align}

Integrating along $u = (1-t, 1-t, 0)^T$ gives $du = (-1, -1, 0)^T dt$ and \begin{align} W_3 &= \int\limits_0^1 (1-t,1-t,*)^T\cdot(-1,-1,0)^T dt \\ &= \int\limits_0^1 (2t-2)dt \\ &= [t^2 - 2t]_0^1 \\ &= -1 \end{align} This would give $$W = W_1 + W_2 + W_3 = \frac{17}{6} - \frac{7}{3} - 1 = -\frac{1}{2}$$

• (+1) In spite of my typo, you've gotten the correct answer! – robjohn Jan 28 '15 at 18:16

If you have $\mathbf r_1(t)=\left\langle t,t,t\right\rangle$ to go from $A$ to $B$, at $t=0$ you are at $\langle 0,0,0\rangle=A$, if $t=1$ you have reached $\langle 1,1,1\rangle =B$. Normally when you construct your parameterizations this way, for example $\mathbf r_2(t)=B+(B-C)t=\langle 1,1,1\rangle + \langle 0,0,-1\rangle t = \langle 1,1,-t\rangle$, you will have the limits $0\le t\le 1$.

• what if finding the limits are not trivial like the case here? – snowman Jan 28 '15 at 13:45
• @snowman There isn't a formula to find limits, you have to think about how you want to parameterize (always a good parameterization with no loops and no going backwards) and then you want a $t$ for which you will be in the first point of the curve, and for which $t$ you will be in the last point of the curve. Then you found the limits. – Vladimir Vargas Jan 28 '15 at 13:48